How does one manually throw a "overflow exception" in C++?

Question

I am rather new to exceptions. In a part of my program if a specific variable goes beyond a certain value I want to manually throw an exception which will say that the variable has gone outside of accepted range. The variable being 16 bits can still hold the "invalid" values but under normal program operation it will only hold a subset of the possible values it can hold.

How do I generate this specific type of exception?

Please ask your question your rubber duck and delete it (not the duck) — , Sep 27 '16 at 21:38
C++ does not have the concept of an "overflow exception". Instead, overflowing integers will silently trigger undefined behavior and/or wrapping. If you want a specific type of exception, you will need to create your own. — Rufflewind, Sep 27 '16 at 21:40
Reopened because OP is not trying to detect an actual overflow, but trying to detect some kind of "logical out-of-range" instead. — Baum mit Augen, Sep 27 '16 at 21:46
well, if I have to "create my own exception" then I shall just look into it then — quantum231, Sep 27 '16 at 21:47
What is wrong here: I have not "marked [this question] as duplicate" !!! — , Sep 27 '16 at 21:47
Dupe candidate would be https://stackoverflow.com/questions/148511/limiting-range-of-value-types-in-c for example. — Baum mit Augen, Sep 27 '16 at 21:48

score 5 · Answer 1 · answered Sep 27 '16 at 21:48

5

throw std::out_of_range("Your value doesn't belong to a valid range");

The class std::out_of_range is defined in

#include <stdexcept>

Documentation: std::out_of_range - cppreference.com

answered Sep 27 '16 at 21:48

ABu

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How does one manually throw a "overflow exception" in C++?

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